Chapter 19, Problem 19.1P

Interpretation Introduction

(a)

Interpretation:

The product of the given reaction is to be predicted.

Concept introduction:

The C atom in the $\text{C}-\text{X (X=halogen)}$ is relatively electron-poor. When an alkyl, alkenyl, or aryl halide reacts with magnesium, the metal atom is inserted between the carbon and the halogen. The C atom is now electron-rich because it is now bonded to a less electronegative metal atom. This reverses the initial polarity (charge) of the carbon, allowing it to form a bond with an electron-poor carbon from say a carbonyl carbon.

Answer to Problem 19.1P

The product of the given reaction is

Explanation of Solution

The given reaction is

Magnesium is an active metal and reacts with the polar $\text{C}-\text{I}$ bond. It extracts the iodine first via homolysis, making the carbon relatively electron-rich. This carbon then forms a bond with the metal atom. Effectively, the Mg atom is inserted between C and I.

Thus, the product of the reaction is phenyl magensium iodide:

Conclusion

The structure of the product was drawn based on reversal of the charge on the carbon atom from the $\text{C}-\text{I}$.

Interpretation Introduction

(b)

Interpretation:

The product of the given reaction is to be predicted.

Concept introduction:

The C atom in the $\text{C}-\text{X (X=halogen)}$ is relatively electron-poor. When an alkyl, alkenyl, or aryl halide reacts with magnesium, the metal atom is inserted between the carbon and the halogen. The C atom is now electron-rich because it is now bonded to a less electronegative metal atom. This reverses the initial polarity (charge) of the carbon, allowing it to form a bond with an electron-poor carbon from say a carbonyl carbon.

Answer to Problem 19.1P

The product of the given reaction is

Explanation of Solution

The given reaction is

The $\text{C}-\text{Br}$ bond is a polar bond, with a partial positive charge on the carbon atom. The metal atom extracts bromine via homolysis. This results in a relatively electron-rich C which then forms a bond with the Mg atom giving the product.

In essence, the reaction inserts the metal atom between the C and Br atoms and leads to a reversal of the charge on the carbon atom.

Thus, the product of the reaction is

Conclusion

The product of the reaction was drawn based on the reversal of charge on the C atom bonded to bromine.

Interpretation Introduction

(c)

Interpretation:

The product of the given reaction is to be predicted.

Concept introduction:

The C atom in the $\text{C}-\text{X (X=halogen)}$ is relatively electron-poor. When an alkyl, alkenyl, or aryl halide reacts with Li(s), the resulting C atom is electron-rich because the C is now bonded to a less electronegative metal atom. This reverses the initial polarity (charge) of the carbon, allowing it to form a bond with an electron-poor carbon from say a carbonyl carbon.

Answer to Problem 19.1P

The product of the reaction is

Explanation of Solution

The given reaction is

Li is a very active metal and will extract the iodine atom from the substrate. The $\text{C}-\text{I}$ bond undergoes homolysis, converting the carbon atom to a relatively electron-rich radical. This radical then reacts with and forms a bond to another Li atom.

Effectively, the reaction results in the replacement of the iodine atom by Li atom, therefore, the product will be

Conclusion

The product of the reaction was drawn based on the reversal of the charge on the carbon atom initially bonded to iodine.

Interpretation Introduction

(d)

Interpretation:

The product of the given reaction is to be drawn.

Concept introduction:

The C atom in the $\text{C}-\text{X (X=halogen)}$ is relatively electron-poor. When an alkyl, alkenyl, or aryl halide reacts with Li(s), the resulting C atom is electron-rich because the C is now bonded to a less electronegative metal atom. The resulting alkyl lithium compound reacts with copper(I) iodide to form a lithium dialkylcuprate.

This reverses the initial polarity (charge) of the carbon, allowing it to form a bond with an electron-poor carbon from say a carbonyl carbon.

Answer to Problem 19.1P

The product of the given reaction is

Explanation of Solution

The given reaction is

The product from part (c) is cyclopentyllithium. Two moles of this compound will react with one of copper(I) iodide to form $\text{C}-\text{Cu}$ bonds.

The carbon atom in the $\text{C}-\text{Cu}$ is partially negatively charged.

The product of the reactio is, therefore,

Conclusion

The product of the reaction was drawn based formation of a bond between the relatively electron-rich carbon atoms and the copper atom.